Question: Let $f(x) = -x^{2}-x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-x^{2}-x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -1, b = -1, c = 8$ $ x = \dfrac{+ 1 \pm \sqrt{(-1)^{2} - 4 \cdot -1 \cdot 8}}{2 \cdot -1}$ $ x = \dfrac{1 \pm \sqrt{33}}{-2}$ $ x = \dfrac{1 \pm \sqrt{33}}{-2}$ $x =\dfrac{1 \pm \sqrt{33}}{-2}$